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very basic doorbell

Hi all, I'm designing a very simple doorbell circuit. I am very new to electronics, so I'd really like to verify my idea with someone. I'll get a 10mA, 12 volt piezo buzzer from Radio Shack and plan on running a 12 volt battery in series with a 1200 ohm resistor to the buzzer from a push button. I'm using 1200 ohm resistor based on R = V / I, that is 12 volts / .01 amps = 1200 ohms. I have a paranoia about starting a fire -- under what conditions, if any, could a circuit like this start a fire? Thanks so much if anyone wants to indulge in such an elementary post....

Reply to
4grpposting
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On a sunny day (Mon, 24 Sep 2007 18:41:43 -0000) it happened snipped-for-privacy@gmail.com wrote in :

That likely means it uses 10mA when it is connected to 12 V. You do not need and do not want the resistor. The thing itself is about 1200Ohm

and plan on

A battery with enough Ah and a low enough Ri together with a short in the buzzer *could* do it. If you are paranoid use a 50mA fuse in series.

Reply to
Jan Panteltje

thanks for the response. I just realized I should have posted in sci.electronics.basics - sorry about that.

Reply to
4grpposting

What you calculated is the total resistance; that is the value of the external resistor, plus the resistance of the buzzer As somebody poined out, the buzzer is probably about 1200 Ohms by itself. That would be how they got the 10 ma.

Tam

Reply to
Tam/WB2TT

cool - got it. One more question, how is a 50mA fuse determined? That is, why a 50mA, not 40, etc.

thanks a lot...

Reply to
4grpposting

The fuse value is not particularly critical - you want it large enough that the fuse will not blow during normal operation, but small enough that a true fault will blow it. You also want a commercially-available value, which may be why 50 mA was suggested, rather than 40 mA.

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Reply to
Peter Bennett

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